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A hammer exerts 49.8 N of force on the head (r=0.00510 m) of a nail. How much pressure does it exert on the nail?

Kisachek [45]

Answer:

609547.12 Pa ≈ 6.10×10^5 Pa

Explanation:

Step 1:

Data obtained from the question. This include the following:

Force (F) = 49.8 N

Radius (r) = 0.00510 m

Pressure (P) =..?

Step 2:

Determination of the area of the head of the nail.

The head of a nail is circular in nature. Therefore, the area is given by:

Area (A) = πr²

With the above formula we can obtain the area as follow:

Radius (r) = 0.00510 m

Area (A) =?

A = πr²

A = π x (0.00510)²

A = 8.17×10^-5 m²

Therefore the area of the head of the nail is 8.17×10^-5 m²

Step 3:

Determination of the pressure exerted by the hammer.

This is illustrated below:

Force (F) = 49.8 N

Area (A) = 8.17×10^-5 m²

Pressure (P) =..?

Pressure (P) = Force (F) /Area (A)

P = F/A

P = 49.8/8.17×10^-5

P = 609547.12 N/m²

Now, we shall convert 609547.12 N/m² to Pa.

1 N/m² = 1 Pa

Therefore, 609547.12 N/m² = 609547.12 Pa.

Therefore, the pressure exerted by the hammer on the nail is 609547.12 Pa or 6.10×10^5 Pa

8 0

4 years ago

One hazard of space travel is debris left by previous missions. There are several thousand objects orbiting Earth that are large

Citrus2011 [14]

Answer:

The correct answer is "6666.67 N".

Explanation:

The given values are:

Mass,

m = 0.100

Relative speed,

v = 4.00 x 10³

time,

t = 6.00 x 10⁻⁸

As we know,

⇒ $F=m(\frac{\Delta v}{\Delta t} )$

On substituting the given values, we get

⇒ $=0.100\times 10^{-6}(\frac{4\times 10^3}{6\times 10^{-8}} )$

⇒ $=6666.67 \ N$

7 0

3 years ago

A certain superconducting magnet in the form of a solenoid of length 0.56 m can generate a magnetic field of 6.5 T in its core w

Gnom [1K]

Answer:

The value is $N =36203 \ turns$

Explanation:

From the question we are told that

The length of the solenoid is $l = 0.56 \ m$

The magnetic field is $B = 6.5 \ T$

The current is $I = 80 \ A$

The desired temperature is $T = 4.2 \ K$

Generally the magnetic field is mathematically represented as

$B = \frac{\mu_o * N * I }{L }$

=> $N = \frac{B * L }{\mu_o * I }$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^{-7} N/A^2$

So

$N = \frac{6.5 * 0.56 }{ 4\pi * 10^{-7} * 80 }$

=> $N =36203 \ turns$

6 0

3 years ago

A neutral copper block is polarized as shown in the figure below, due to an electric field made by external charges (notshown).

Fudgin [204]

Answer:

As point B is located inside the copper block so net electric field at point B is j.

Explanation:

Consider the figure attached below. The net electric field at location B,that is inside the copper block is zero because when a conductor is charged or placed in an electric field of external charges, net charge lies on the surface of conductor and there is no electric field inside the conductor. As point B is located inside the copper block so net electric field at point B is zero as well direction of net electric field at point B is zero.

6 0

3 years ago

What minimum speed does a 200 g puck need to make it to the top of a frictionless ramp that is 4.1 m long and inclined at 22 ∘?

myrzilka [38]

Answer:

5.5 m/ sec

Explanation:

Because the inclined surface is frictionless so we can assume that total change of energy is zero

i-e ΔE = 0

Or we can say that difference between final and initial energy is zero i-e

Ef- Ei =0

Where,

Ef= final energy at the top of the ramp= KEf+PEf

Ei= Initial energy at the bottom of the ramp=KEi+PEi

So we have

(KEf+PEf)-(KEi+PEi)=0

==>KEf-KEi+PEf-PEi=0 -------------(1)

KEf = mgh = 200×9.8×h

Where h= Sin 22 = h/d= h/4.1

or

0.375×4.1=h

or h= 1.54 m

So, PEf= 200×9.8×1.54=3018.4 j

and KEf= 1/2 m $Vf^{2}$ = 0.5×200×0=0 j

PEi= mgh = 200×9.8×0=0 j

KEi= 1/2 m $Vi^{2}$ =0.5×200× $Vi^{2}$ =100 $Vi^{2}$ j

Put these values in eq 1, we get;

0-100 $Vi^{2}$ +3018.4-0=0

-100 $Vi^{2}$ =-3018.4

==> $Vi^{2}= \frac{3018.4}{100}$ = 30.184

==> Vi = $\sqrt{30.184} = 5.5 m.sec$

7 0

3 years ago